If a = lim x ? 0 1 + 1 + x 4 - 2 x 4 and b = lim x ? 0 sin 2 x 2 - 1 + cos x , then the value of a b 3 is: [2024]

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Solution

Q.
$If a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} and b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}, then the value of a b^{3} is: [2024]$

1 25

2 36

3 32

4 30

Ans.
(3)

$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}}$

Rationalising numerator, we get

$a = \lim_{x \to 0} \frac{1 + \sqrt{1 + x^{4}} - 2}{x^{4} (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})}$

$= \lim_{x \to 0} \frac{\sqrt{1 + x^{4}} - 1}{x^{4} (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})}$

Again rationalising numerator, we get

$= \lim_{x \to 0} \frac{1}{(\sqrt{1 + x^{4}} + 1) (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})} = \frac{1}{2 \times 2 \sqrt{2}} = \frac{1}{4 \sqrt{2}}$

Now, $b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}$

$= \lim_{x \to 0} \frac{(1 - \cos^{2} x) (\sqrt{2} + \sqrt{1 + \cos x})}{[2 - (1 + \cos x)]}$ [By rationalising denominator]

$= \lim_{x \to 0} \frac{(1 + \cos x) (1 - \cos x) (\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$

$= \lim_{x \to 0} (1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x}) = 4 \sqrt{2}$

$∴$ $a b^{3} = \frac{1}{4 \sqrt{2}} \times {(4 \sqrt{2})}^{3} = 32$

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